200k views
1 vote
In a hydrogen atom (i.e., one stationary proton and one orbiting electron), the angular velocity of the electron is 10 x 10^6 radians per second. What is the radius of the electron orbit? Assume uniform circular motion.

1 Answer

3 votes

Assuming the centripetal force is provided by the Coulomb attraction between the electron and the proton we have that,


F_c = k (e^2)/(R^2)

Here,

k = Coulomb's Constant

R = Distance

e = Electron charge

And by the centripetal force,


F_c = m_e R\omega^2


m_e = Mass of electron

R = Radius


\omega = Angular velocity

Equation both expression,


k (e^2)/(R^2) = m_e R\omega^2

Replacing,


R^3 = ((9*10^9)(1.6*10^(-19))^2)/((9.11*10^(-31))(10^(6))^2)


R^3 = 2.52908*10^(-10) m


R = 6.32394*10^(-4)m

Therefore the radius of the electron orbit is
6.32394*10^(-4)m

User Carrabino
by
7.5k points